6 Answers
6

For a linearly reductive (i.e. all representations are completely redudible) group G, any quotient G/H is also linearly reductive. Given a representation V of G/H, you get an induced representation of G (with the same underlying vector space). Then any decomposition of V as a representation of G is also a decomposition as a representation of G/H. Since G-invariant subspaces of V are the same thing as G/H-invariant subspace of V, the notions of irreducible subrepresentations agree. So you get a complete decomposition of V as a representation of G/H.

In characteristic zero, reductive is equivalent to linearly reductive, so you have your answer. In general, a group is reductive if its unipotent radical is trivial. I feel like you should be able to argue that an element of the unipotent radical of G/H lifts to give you an element of the unipotent radical of G, but I'm not sure.

It is important in answering this question that one can extend scalars to a perfect (e.g., algebraically closed) ground field, as was implicit in many of the other answers even if not said explicitly. Indeed, if $k$ is an imperfect field then it always happens that there exist many examples of pairs $(G,H)$ with $G$ a smooth connected affine $k$-group containing no nontrivial smooth connected unipotent normal $k$-subgroup and $H$ a smooth connected normal $k$-subgroup in $G$ such that $G/H$ contains a non-trivial smooth connected unipotent normal $k$-subgroup.
This can even happen when $G$ and $H$ are perfect (i.e., equal to their own derived groups), which is really disorienting if one is accustomed to working with reductive groups.

For a simple commutative example, let $k'/k$ be a purely inseparable extension of degree $p = {\rm{char}}(k)$ and let $G$ be the Weil restriction ${\rm{Res}} _{k'/k}(\mathbf{G} _m)$, which is just "${k'}^{\times}$ viewed as a $k$-group". This contains a natural copy of $\mathbf{G}_m$, and since $k'/k$ is purely inseparable this $k$-subgroup $H$ is the unique maximal $k$-torus and the quotient $G/H$ is unipotent of dimension $p-1$ (over $\overline{k}$ it is a power of $\mathbf{G} _a$ via truncation of $\log(1+x)$ in degrees $< p$, as one sees using the structure of the $\overline{k}$-algebra $\overline{k} \otimes_k k'$). The main point then is that $G$ itself contains no nontrivial smooth connected unipotent $k$-subgroups, which is true because we are in characteristic $p > 0$ and $G$ is commutative with $G(k_s)[p] = {k'_s}^{\times}[p] = 1$! Note: the unipotent quotient $G/H$ is an example of a smooth connected unipotent $k$-group (even commutative and $p$-torsion) which contains no $\mathbf{G}_a$ as a $k$-subgroup (proof: commutative extensions of $\mathbf{G}_a$ by $\mathbf{G}_m$ split over any field, due to the structure of ${\rm{Pic}}(\mathbf{G}_a)$ and a small calculation); that is, $G/H$ is a "twisted form" of a (nonzero) vector group, which never happens over perfect fields.

Making examples with perfect $G$ and $H$ is less straightforward; see Example 1.6.4 in the book "Pseudo-reductive groups".

As for the suggestion to use Haboush's theorem (whose proof I have never read), I wonder if that is circular; it is hard to imagine getting very far into the theory of reductive groups (certainly to the point of proving Haboush's theorem) without needing to already know that reductivity is preserved under quotients (a fact that is far more elementary than Haboush's theorem, so at the very least it seems like killing a fly with a sledgehammer even if it is not circular).

Finally, since nobody else has mentioned it, look in any textbook on linear algebraic groups (Borel, Springer, etc.) for a proof of the affirmative answer to the original question. For example, 14.11 in Borel's book. Equally important in the theory is that for arbitrary smooth connected affine groups, formation of images also commutes with formation of maximal tori and especially (scheme-theoretic) torus centralizers; see corollaries to 11.14 in Borel's book.

Allow me to insert a trivial word of caution: If analytic quotients are allowed, then the assertion is false even in characteristic zero, as the expression $$\mathbb{G}_m/q^{\mathbb{Z}}$$ for an elliptic curve will show.

Well, that case can be included. But over $\mathbb{C}$, any elliptic curve can be written this way.
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Minhyong KimJan 9 '10 at 0:01

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Is it clear than an elliptic curve is non-reductive? It seems to be both reductive -- since it does not contain any nontrivial connected normal unipotent subgroup, and linearly reductive -- it admits no nontrivial linear representations, so all of its representations are completely reducible. Or is the criterion of linearity explicitly part of the definition of a reductive group?
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Pete L. ClarkJan 9 '10 at 0:01

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@Pete Clark: By the way, the domain of the adjective' you introduce is no reason to admit the ambiguity you propose. I can safely say I am not a reductive group,' as I'm inclined to say an abelian variety is not a reductive group.
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Minhyong KimJan 9 '10 at 10:36

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In defense of Wikipedia, it does say that the algebraic group must be affine (in other words linear) to begin with.
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Wilberd van der KallenMar 28 '10 at 14:47

Yes, I believe the quotient of a reductive group scheme is reductive. Perhaps it's possible to show directly as Anton was suggesting. I have an argument that shows that if G is geometrically reductive then G/H is geometrically reductive. The statement that G reductive implies G/H reductive follows from Haboush's theorem (ie. G is reductive if and only if G is geometrically reductive).

Geometrically reductive means that if G acts algebraically on an algebra A and I is an invariant ideal, then any element of A/I that is fixed by G has a power that lifts to an element of A that is fixed by G. There are other formulations. It is now easy to see that a quotient of a geometrically reductive G is geometrically reductive. One does not need Haboush's theorem, which is much deeper.
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Wilberd van der KallenMar 28 '10 at 14:37

Of course Brian Conrad is right that invoking Haboush's theorem is like killing a fly with a sledgehammer. But I like this particular use of a sledgehammer. It reduces the problem to one where one can argue as in the characteristic zero case. It makes one feel the result is true for a reason.
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Wilberd van der KallenMar 28 '10 at 17:25

Presumably the argument ilya was driving at is this: If one had a surjective map of group schemes $G \to H$, then consider the preimage of the nilradical of $H$. This is an algebraic group with a surjective map to a unipotent group. Since there are no group homomorphisms from reductive groups to unipotent ones, the nilradical of this preimage (which is contained in the nilradical of $G$) must surject onto the nilradical of $H$. So if the former is trivial, so is the latter.